Slove A. 115°B. 58°C. 50°D. 226°

Mathematics

2 answers:

lana66690 [7]3 years ago

7 0

The answer is C on edgenuity.

Diano4ka-milaya [45]3 years ago

6 0

Answer:

Step-by-step explanation:

The angle is total is 138 degrees. Part of the entire angle is already given (the 88 degree measure), so all you have to do is subtract 88 from 138, which is 50.

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Last week, the price of cherries at the corner deli was $4.99 per pound. This week, cherries at the same deli are on sale at a 1

sergij07 [2.7K]

Answer:

The price for 3.75 pounds of cherries is $17.775.

Step-by-step explanation:

The price las week was $ 4.99 and it got a discount of 15% for this week, this means that the price for this week is 100% of the one prior minus 15%, therefore it is 85% of the last price. We need to calculate the price per pound with this discount as shown below:

$\text{this week} = \text{last week}*\frac{95}{100}\\\\\text{this week} = 4.99*0.95 = 4.74 \text{ per pound}\\$

Since the price is $4.74 per pound and we want to buy 3.75 pounds, then the total amount we need to pay is:

$\text{total amount} = 3.75*4.74 = 17.775$

The price for 3.75 pounds of cherries is $17.775.

6 0

3 years ago

(cotx+cscx)/(sinx+tanx)

Butoxors [25]

Answer: $\bold{\dfrac{cot(x)}{sin(x)}}$

<u>Step-by-step explanation:</u>

Convert everything to "sin" and "cos" and then cancel out the common factors.

$\dfrac{cot(x)+csc(x)}{sin(x)+tan(x)}\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)}{1}+\dfrac{sin(x)}{cos(x)}\bigg)\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg[\dfrac{sin(x)}{1}\bigg(\dfrac{cos(x)}{cos(x)}\bigg)+\dfrac{sin(x)}{cos(x)}\bigg]\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)}{cos(x)}+\dfrac{sin(x)}{cos(x)}\bigg)$

$\text{Simplify:}\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)+sin(x)}{cos(x)}\bigg)\\\\\\\text{Multiply by the reciprocal (fraction rules)}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)cos(x)+sin(x)}\bigg)\\\\\\\text{Factor out the common term on the right side denominator}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)(cos(x)+1)}\bigg)$

$\text{Cross out the common factor of (cos(x) + 1) from the top and bottom}:\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)}\bigg)\\\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times cot(x)}\qquad \rightarrow \qquad \dfrac{cot(x)}{sin(x)}$